1. **Problem Statement:** We are given the Boolean function $F(A,B,C) = \Sigma(0,2,3,6)$, which means $F$ is true for minterms 0, 2, 3, and 6. 2. **Create the Truth Table and Minterms:** The minterms correspond to the binary values of $A,B,C$ where $F=1$. | A | B | C | F | |---|---|---|---| | 0 | 0 | 0 | 1 | (m_0 = \overline{A}\overline{B}\overline{C}) | 0 | 0 | 1 | 0 | | 0 | 1 | 0 | 1 | (m_2 = \overline{A}B\overline{C}) | 0 | 1 | 1 | 1 | (m_3 = \overline{A}BC) | 1 | 0 | 0 | 0 | | 1 | 0 | 1 | 0 | | 1 | 1 | 0 | 1 | (m_6 = A B \overline{C}) | 1 | 1 | 1 | 0 | 3. **Sum of Product Terms (SOP):** $$F = \overline{A}\overline{B}\overline{C} + \overline{A}B\overline{C} + \overline{A}BC + AB\overline{C}$$ 4. **Simplify Using Boolean Algebra:** Group terms: - Group 1: $\overline{A}\overline{B}\overline{C} + \overline{A}B\overline{C} = \overline{A}\overline{C}(\overline{B} + B) = \overline{A}\overline{C}$ (since $\overline{B} + B = 1$) - Group 2: $\overline{A}BC + AB\overline{C}$ cannot be combined directly, but check further: Rewrite $\overline{A}BC + AB\overline{C}$ as $B(\overline{A}C + A\overline{C})$ which is $B(A \oplus C)$. So, $$F = \overline{A}\overline{C} + B(A \oplus C)$$ This is a simplified expression using XOR, but since only AND, OR, NOT gates are allowed, express $A \oplus C$ as: $$A \oplus C = A\overline{C} + \overline{A}C$$ So, $$F = \overline{A}\overline{C} + B(A\overline{C} + \overline{A}C)$$ 5. **Create Truth Table for Simplified Expression:** Check $F = \overline{A}\overline{C} + B(A\overline{C} + \overline{A}C)$ matches original: | A | B | C | $\overline{A}\overline{C}$ | $A\overline{C}$ | $\overline{A}C$ | $B(A\overline{C} + \overline{A}C)$ | $F$ | |---|---|---|----------------|--------------|-------------|-----------------------------|-----| | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | Matches original truth table. 6. **Logic Diagram:** - Use NOT gates to get $\overline{A}$ and $\overline{C}$. - Use AND gates for $\overline{A}\overline{C}$, $A\overline{C}$, and $\overline{A}C$. - Use OR gates to combine $A\overline{C} + \overline{A}C$ and then OR with $\overline{A}\overline{C}$. - Finally, AND with $B$ for the middle term. This completes the first problem.